a) A 1.00 μL sample of an equal volume mixture of 2-pentanone and 1-nitropropane is injected into a gas chromatograph. T

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a) A 1.00 μL sample of an equal volume mixture of 2-pentanone and 1-nitropropane is injected into a gas chromatograph. T

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a) A 1.00 μL sample of an equal volume mixture of 2-pentanone and 1-nitropropane is injected into a gas chromatograph. The densities of these compounds are 0.8124 g/mL for 2-pentanone and 1.0221 g/mL for 1-nitropropane. What mass of each compound was injected? Mass of 2-pentanone = ✔0.406 = 4.0 0.511 Mass of 1-nitropropane mg b) The peak areas produced on this injection were 1562 units for 2-pentanone and 1258 units for 1-nitropropane. Calculate the response factor for each compound as area per mg. 2-pentanone: 4.0 3850 units/mg units/mg 1-nitropropane: 4.0 2460 c) An unknown mixture of these two components produces peak areas of 1067 units (2-pentanone) and 1264 units (1-nitropropane). 2-pentanone: 4.0 40.6 1-nitropropane: 4.0 51.1 mg ✔ Use these areas and the response factors above to determine the weight % of the components in the unknown sample. X % X %
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